I just can't see it. I would think that it would involve some form of trig substitution, but I'm just drawing a blank. I'll do the work if someone can please give me a nice little hint. I know that [tex]\int \arcsin{(x)}dx = \sqrt{1-x^2}+ x\arcsin{(x}) + C[/tex] from my calculator and mathematica. Hint please. :yuck:
If only you were working with its derivative! I bet you know how to integrate something that looks like its derivative!
Sure... I'd draw a nice triangle and do trig substitution if it was [tex]\int \frac{dx}{\sqrt{1-x^2}}[/tex] Hmmm...I'll think out loud here. If the integral has a square-root and is in the form of [itex]c^2-x^2[/itex] then x is one leg, c is the hypotenuse, and the other leg is the previously mentioned radical. Is trig substitution with right triangles on the right track? Since the hypotenuse is [itex]\sqrt{a^2+b^2}[/itex], it seems that one leg might need to be [itex]\sqrt{\arcsin{(x)}}[/itex] Somehow I don't think I'm on track.
well, you know the integral of sinx with limits. Now arcsin x will be the limits, and you can make a rectangle.
Or you could just take the derivative of the right hand side and go "ta da!" and that's proof enough for me.
hint equate the arc sine to another variable e.g y.making it a sine fxn.e.g the arc sine of 0.5=30,while sine30=0.5.this will simplify the integral and further substitution will conclude it
Just for the fun of it ... The sum of the integrals [tex]\int \sin^{-1} x dx + \int \sin y dy[/tex] is just the area of the bounding rectangle: [itex] x \times \sin^{-1} x[/itex] Since [tex]\int \sin y dy = -\cos y + C[/tex] and [tex]\cos y = \cos \sin^{-1} x = \sqrt {1-x^2}[/tex] it follows that [tex]\int \sin^{-1} x dx = \sqrt {1-x^2} + x \sin^{-1} x + C[/tex]
Integration by parts? Hey, Im really sorry to arise dead threads from the past (which i have seen though google) but somehting really weird happened me when I tried to use integration by parts on arcsinx. let me show you: S(arcsinx)= {v'(x)=1} {u(x)=arcsinx} xarcsinx-S(x*d(arcsinx))= xarcsinx-S(x/(1-x^2)^0.5= {u(x)=x v'(x)=arcsinx} xarcsinx-xarcsinx+S(arcsin)dx ==> S(arcsinx)=S(arcsinx) :\ I know I have done something really stupid here, but please be easy on me since I started studying Integrals only three days ago. In the second time I used integration by parts, do I miss something , is there another efficient choice of v and u? Thanks in advance, Aviv p.s: I will edit it better to use normal math signs once I figure out how.
Im sorry, Tried it also and all i got is: xarcsine x -(x^2/2)(1/(Sqrt(1-x^2)))-(x/4)(sqrt(1/(1-x^2))+1/4(arcsinx) this isn't going anywhere :(
Make [itex] x=\sin t [/itex]. Then apply part integration on the resulting integral. It's just a way to avoid the simple solution of part integrating directly.
solve it officially. Fixed during some major mistakes I had about dev' and stuff. did it without subtition, only using integration by parts. if you are interested what I did then you are welcome to tell me to write my solution. Thanks guys :) gg
use seperation by parts u=arcsinx du=dx/(1-x^2)^1/2 dv=dx v=x uv-ingegral vdu = xarsinx-integral x/(1-x^2)^1/2 use u substitution with u = 1-x^2 so du = -2x than you get xarsinx + (1-x^2)^1/2 + c
f(x)=arcsinx f'(x)=1/radical(1-xsquare) g'(x)=1 g(x)=x // S means integral S arcsin x dx= x arcsin x - S xdx/radical(1-xsquare) = x arcsin x + S (radical(1-xsquare))'dx= =x arcsin x + radical(1-xsquare) +C